For each polynomial $p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ we define it's derivative as this and we show it by $p'(x)$: \[p'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+...+2a_2x+a_1\] a) For each two polynomials $p(x)$ and $q(x)$ prove that:(3 points) \[(p(x)q(x))'=p'(x)q(x)+p(x)q'(x)\] b) Suppose that $p(x)$ is a polynomial with degree $n$ and $x_1,x_2,...,x_n$ are it's zeros. prove that:(3 points) \[\frac{p'(x)}{p(x)}=\sum_{i=1}^{n}\frac{1}{x-x_i}\] c) $p(x)$ is a monic polynomial with degree $n$ and $z_1,z_2,...,z_n$ are it's zeros such that: \[|z_1|=1, \quad \forall i\in\{2,..,n\}:|z_i|\le1\] Prove that $p'(x)$ has at least one zero in the disc with length one with the center $z_1$ in complex plane. (disc with length one with the center $z_1$ in complex plane: $D=\{z \in \mathbb C: |z-z_1|\le1\}$)(20 points)
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Tags: algebra, polynomial, calculus, derivative, algebra solved
09.09.2010 14:13
Is there any solutions?
11.09.2010 19:47
I've asked my teacher to send me the solution, do you want me to post it?
12.09.2010 09:21
goodar2006 wrote: I've asked my teacher to send me the solution, do you want me to post it? Yes. Thank you very much! I'm very confused about this question!!
13.09.2010 08:46
WLOG you can assume that $z_1=1$ (why?) so $p(x)=(x-1)q(x)$. suppose that the problem is not true, and define $R(x)=p'(x+1)$, so $R(x)$ doesn't have any root's in the disc with length $1$ and it's center on the origin of coordinates. now we have $|\frac{q'(1)}{q(1)}|=|\frac{R'(0)}{2R(0)}|< \frac{n-1}{2}$ but we also have $Re(\frac{q'(1)}{q(1)})\ge \frac{n-1}{2}$, a contradiction.
23.10.2010 11:39
goodar2006 wrote: now we have $|\frac{q'(1)}{q(1)}|=|\frac{R'(0)}{2R(0)}|< \frac{n-1}{2}$ but we also have $Re(\frac{q'(1)}{q(1)})\ge \frac{n-1}{2}$. Hello my friend, I have been busy in preparing a very important exam for Chinese students in Oct 17 and didn't read your answer carefully before. Could you explain how could we get this? Thanks.
12.11.2010 14:13
litongyang wrote: goodar2006 wrote: now we have $|\frac{q'(1)}{q(1)}|=|\frac{R'(0)}{2R(0)}|< \frac{n-1}{2}$ but we also have $Re(\frac{q'(1)}{q(1)})\ge \frac{n-1}{2}$. Hello my friend, I have been busy in preparing a very important exam for Chinese students in Oct 17 and didn't read your answer carefully before. Could you explain how could we get this? Thanks. because R(x) has no root in {z∈C| |z|<=1} so we have |R'(x)/R(x)| = |1/x1 +1/x2 + ...+1/xn-1|< n-1 Re(q'(1)/q(1))=Re(1/1-Z2) + Re(1/1-Z3) +....+ Re(1/1-Zn), note that 2Re(1/1-x)=(2-2Rex)/1-2Rex+|x||x|,and|x|<=1, without much caculate,we can know that when x=-1,Re(1/1-x) get the min value,so Re(q'(1)/q(1))>=(n-1)/2 that get our result
12.11.2010 16:28
goodar2006 wrote: b) Suppose that $p(x)$ is a polynomial with degree $n$ and $x_1,x_2,...,x_n$ are it's zeros. prove that:(3 points) \[\frac{p'(x)}{p(x)}=\sum_{i=1}^{n}\frac{1}{x-x_i}\] we have: $p(x)=a_n(x-x_1)(x-x_2)...(x-x_n)$ $\Rightarrow p'(x)=a_n[ (x-x_2)(x-x_3)...(x-x_n)+...+(x-x_1)(x-x_2)...(x-x_{n-1})=\sum^{n}_{i=1} \Pi^{n}_{j=1, j\neq i}(x-x_j)$ $\Rightarrow \frac{p'(x)}{p(x)}=\sum_{i=1}^{n}\frac{1}{x-x_i}$ $\Rightarrow Q.E.D$